将ks.test，var.test，t.test和wilcox.test组合成一个类似于决策树的函数或者如果是函数在r中

Question

我的数据如下：

df1 <- read.table(text = "A1 A2 A3 A4 B1 B2 B3 B4
1 2 4 12 33 17 77 69
34 20 59 21 90 20 43 44
11 16 23 24 19 12 55 98
29 111 335 34 61 88 110 320
51 58 45 39 55 87 55 89", stringsAsFactors = FALSE, header = TRUE, row.names=c("N1","N2","N3","N4","N5"))

我想逐行比较A和B之间的值。首先，我想测试A和B的分布是否由ks.test正态分布。其次，我将测试A和B之间的差异是否因var.test而异。对于非正态分布结果（p ks.test <0.05），我将通过wilcox.test进行wilcox测试。对于正态分布式结果，我将通过t.test将它们分成相等和不等方差ttest来进行ttest。最后，我结合了所有结果。

我所做的是，首先，我设置了ks.test，var.test，wilcox.test和两个t.test的五个函数：

kstest<-function(df, grp1, grp2) {
  x = df[grp1]
  y = df[grp2]
  x = as.numeric(x)
  y = as.numeric(y)  
  results = ks.test(x,y,alternative = c("two.sided"))
  results$p.value
}
vartest<-function(df, grp1, grp2) {
  x = df[grp1]
  y = df[grp2]
  x = as.numeric(x)
  y = as.numeric(y)  
  results = var.test(x,y,alternative = c("two.sided"))
  results$p.value
}
wilcox<-function(df, grp1, grp2) {
  x = df[grp1]
  y = df[grp2]
  x = as.numeric(x)
  y = as.numeric(y)  
  results = wilcox.test(x,y,alternative = c("two.sided"))
  results$p.value
}
ttest_equal<-function(df, grp1, grp2) {
  x = df[grp1]
  y = df[grp2]
  x = as.numeric(x)
  y = as.numeric(y)  
  results = t.test(x,y,alternative = c("two.sided"),var.equal = TRUE)
  results$p.value
}

ttest_unequal<-function(df, grp1, grp2) {
  x = df[grp1]
  y = df[grp2]
  x = as.numeric(x)
  y = as.numeric(y)  
  results = t.test(x,y,alternative = c("two.sided"),var.equal = FALSE)
  results$p.value
}

然后我计算了ks.test和var.test的p值来对数据进行子集化：

ks_AB<-apply(df1,1,kstest,grp1=grepl("^A",colnames(df1)),grp2=grepl("^B",colnames(df1)))

ks_AB
[1] 0.02857143 0.69937420 0.77142857 0.77142857 0.21055163

var_AB<-apply(df1,1,vartest,grp1=grepl("^A",colnames(df1)),grp2=grepl("^B",colnames(df1)))

var_AB
[1] 0.01700168 0.45132827 0.01224175 0.76109048 0.19561742

df1$ks_AB<-ks_AB
df1$var_AB<-var_AB

然后我通过上面描述的数据对数据进行子集化：

df_wilcox<-df1[df1$ks_AB<0.05,]
df_ttest_equal<-df1[df1$ks_AB>=0.05 & df1$var_AB>=0.05,]
df_ttest_unequal<-df1[df1$ks_AB>=0.05 & df1$var_AB<0.05,]

最后，我计算了对新数据帧的相应测试，并合并了结果

wilcox_AB<-as.matrix(apply(df_wilcox,1,wilcox,grp1=grepl("^A",colnames(df_wilcox)),grp2=grepl("^B",colnames(df_wilcox))))

ttest_equal_AB<-as.matrix(apply(df_ttest_equal,1,ttest_equal,grp1=grepl("^A",colnames(df_ttest_equal)),grp2=grepl("^B",colnames(df_ttest_equal))))

ttest_unequal_AB<-as.matrix(apply(df_ttest_unequal,1,ttest_unequal,grp1=grepl("^A",colnames(df_ttest_unequal)),grp2=grepl("^B",colnames(df_ttest_unequal))))

p_value<-rbind(wilcox_AB,ttest_equal_AB,ttest_unequal_AB)
colnames(p_value)<-c("pvalue")

df<-merge(df1,p_value,by="row.names")

df
  Row.names A1  A2  A3 A4 B1 B2  B3  B4      ks_AB     var_AB     pvalue
1        N1  1   2   4 12 33 17  77  69 0.02857143 0.01700168 0.02857143
2        N2 34  20  59 21 90 20  43  44 0.69937420 0.45132827 0.39648631
3        N3 11  16  23 24 19 12  55  98 0.77142857 0.01224175 0.25822839
4        N4 29 111 335 34 61 88 110 320 0.77142857 0.76109048 0.85703939
5        N5 51  58  45 39 55 87  55  89 0.21055163 0.19561742 0.06610608

我知道我的代码很乏味和愚蠢，但它对我的数据非常有用。我现在想知道我是否将上面的代码组合到if else函数的新的类似决策树的函数中，它将像：

Answer 1

这应该做的伎俩：

FOO <- function(df, grp1, grp2){

  # perform rowwise kolmogorov smirnov test
  ks_AB <- apply(df, 1, kstest, grp1 = grp1, grp2 = grp2)

  # subset data by significance of ks test
  sub1 <- df[ks_AB < .05, ]
  sub2 <- df[ks_AB >= .05, ]

  if(nrow(sub1) > 0){
    # perform wilcoxon rank sum test on non-normally distributed data
    wilc_AB <- apply(sub1, 1, wilcox, grp1 = grp1, grp2 = grp2)
  }

  if(nrow(sub2) > 0){
    # perform f test on normally distributed data
    var_AB <- apply(sub2, 1, vartest, grp1 = grp1, grp2 = grp2)

    # subset data by significance of f test
    varsub1 <- sub2[var_AB < .05, ]
    varsub2 <- sub2[var_AB >= .05, ]

    if(nrow(varsub1) > 0){
      # perform t test with unequal variance on subset with unequal variance
      t_uneq_AB <- apply(varsub1, 1, ttest_unequal, grp1 = grp1, grp2 = grp2)
    }

    if(nrow(varsub2) > 0){
      # perform t test with equal variance on subset with equal variance
      t_eq_AB <- apply(varsub2, 1, ttest_equal, grp1 = grp1, grp2 = grp2)
    }
  }

  # put together output dataframe
  df$ks_AB <- ks_AB

  if(exists("var_AB")){
    df$var_AB <- NA
    df$var_AB[row.names(df) %in% names(var_AB)] <- var_AB
  }

  df$pvalue <- NA

  if(exists("wilc_AB")){
    df$pvalue[row.names(df) %in% names(wilc_AB)] <- wilc_AB
  }

  if(exists("t_uneq_AB")){
    df$pvalue[row.names(df) %in% names(t_uneq_AB)] <- t_uneq_AB
  }

  if(exists("t_eq_AB")){
    df$pvalue[row.names(df) %in% names(t_eq_AB)] <- t_eq_AB
  }

  # return output
  return(df)
}

应用于您的数据示例：

> FOO(df1, grepl("^A",colnames(df1)), grp2=grepl("^B",colnames(df1)))
   A1  A2  A3 A4 B1 B2  B3  B4      ks_AB     var_AB     pvalue
N1  1   2   4 12 33 17  77  69 0.02857143         NA 0.02857143
N2 34  20  59 21 90 20  43  44 0.69937420 0.45132827 0.39648631
N3 11  16  23 24 19 12  55  98 0.77142857 0.01224175 0.25822839
N4 29 111 335 34 61 88 110 320 0.77142857 0.76109048 0.85703939
N5 51  58  45 39 55 87  55  89 0.21055163 0.19561742 0.06610608
Warning messages:
1: In ks.test(x, y, alternative = c("two.sided")) :
  cannot compute exact p-value with ties
2: In ks.test(x, y, alternative = c("two.sided")) :
  cannot compute exact p-value with ties

您可以在函数中使用suppressWarnings()来抑制这些警告，但我宁愿让它们显示出来，这样您就知道测试可能不准确。